Stiffness of Cracked Rc Structure

Predictingofthestiffnessofcrackedreinforcedconcretestructure YongzhenLi

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FinalReport

Predicting of the Stiffness of Cracked Reinforced

Concrete Structures Author: Yongzhen Li 1531344 Delft University of Technology Faculty of Civil Engineering & Geosciences Department Design and Construction Section Structural and Building Engineering Stevinweg 1, Delft Commissioner: Van Hattum en Blankevoort Korenmolenlaan 2, Woerden Supervisors: Prof. dr. ir. J.C. Walraven TU Delft Prof.dr.ir. Ningxu Han Van Hattum en Blankevoort Dr.ir.drs. C.R. Braam TU Delft Dr.ir. P.C.J. Hoogenboom TU Delft Ir. L.J.M. Houben TU Delft July 2010

PredictingoftheStiffnessofCrackedReinforcedConcreteStructure YongzhenLi

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Acknowledgements This research was conducted at the Faculty of Civil Engineering and Geosciences at Delft University of Technology and Van Hattum en Blankevoort. I would like to thank Prof.dr.ir. J.C. Walraven for his help and encouragement during the past year. I especially would like to thank Dr.ir.drs. C.R. Braam and Prof.dr.ir. Ningxu Han for their valuable guidance and their help throughout this project, without which this work would not have been possible. I also would like to thank Dr.ir. P.C.J. Hoogenboom for giving me a lot of advises and help in solving problems. I would like to thank my coordinator Ir. L.J.M. Houben who helped me a lot in the graduation process. I also would like to thank to Van Hattum en Blankevoort, since they gave me the chance to carry out my thesis work there with a lot of help and advices. Finally, I wish to thank my family for their support and care.


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Summary Cracking is inherent in design of reinforced concrete, and it influences the structures durability and its appearance. If the cracks are too wide, the structure might not fulfill requirements with regard to durability and serviceability e.g. liquid tightness. Therefore a good design and detailing of a structure should be made to limit crack widths. But unexpected cracking might occur. Many factors influence the cracking behavior of concrete structures: Cracks can not only be caused by imposed loads, but also by (partially) restrained imposed deformations. In the latter case there is an interaction between the forces generated and the stiffness of the structure, which is influenced by the cracking behavior: the more the stiffness is reduced by cracking, the lower the forces. It is difficult to make a design in which all influencing factors are taken into account. So, when structural modeling imposed deformations, engineers often reduce the uncracked stiffness when modeling the structure and designing the reinforcement. The question arises which reduction factor to use. In practice, Youngs modulus is often reduced to 1/3 of its original value. Answering the question whether this is a suitable value is the main goal of the research. The research focused on basic theories on cracking behavior. The tension stiffening law is used and it is researched from micro size to macro size, from cross-section to system, from effect to action. Finally, an appropriate stiffness reduction value is obtained. The procedure is: 1. By using the cross section stress balance, the accurate compression zone height will be

obtained under both axial force and bending moment. 2. The elastic modulus is an important parameter related to the moment caused by restrained

deformation. After the compression zone height is obtained, by using the Tension Stiffening Law, the Elastic Modulus in the crack is calculated.

3. After transferring the cross sectional stiffness into system stiffness, the accurate moment curvature curve and the design mean stiffness are obtained.

The design mean stiffness is not constant for different loading combinations. It is larger than one third of the uncracked stiffness when there is a tensile axial force and a high positive temperature gradient. On the other hand, the design mean stiffness might also be less than one third of uncracked stiffness. There is no difference for the loading sequence. That means that whether external loading or restrained deformation is applied first, the results will be same at the final state. After cracking, the non-linear response of the member investigated will influence the bending moment distribution. As a result, the bending moment in a cross-section is not only influenced by external loading and restrained deformation, but also by the stiffness distribution over the length of the member. It is not suitable for engineers to always use one third of the uncracked stiffness to design the reinforcement since they might then underestimate the forces caused by the temperature gradient: It will be higher when there is an axial tensile force in combination with a high positive temperature gradient. A program to obtain the accurate value of the stiffness of a clamped beam is developed. This will help engineers to prepare a more accurate structural model.


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CONTENTS 1.General introduction.....................................................................................................1

1.1 Introduction.................................................................................................................21.2 Problem description....................................................................................................21.3 Goal of the research....................................................................................................31.4 Research outline..........................................................................................................4

2.Literatures survey...........................................................................................................52.1 Different codes in calculation crack...........................................................................6

2.1.1 Crack width calculation equations and comparison....................................62.1.2 Steel stress equations under crack width control and comparison...........122.1.3 Conclusion......................................................................................................13

2.2 General literatures....................................................................................................142.2.1 Introduction of crack width control............................................................142.2.2 Causes of cracks............................................................................................142.2.2.1 External Loading...........................................................................................142.2.2.2 Imposed strain...............................................................................................162.2.3 Loading combination....................................................................................16

3.Calculation of compression zone height.........................................................193.1 H under only tension reinforcement and only N....................................................213.2 H under both compression and tension reinforcement with only M....................233.3 H under only tension reinforcement with both M and N.......................................263.4 H under both compression and tension reinforcement with both M and N.........283.5 The compression zone height equation of Noakowski............................................323.6 Example by using two method of calculation compression zone height...............34

4.Calculation of the stiffness...................................................................................364.1 Cracking Force..........................................................................................................384.2 Bending stiffness in a crack......................................................................................394.3 Difference of the centroidal axis x after moved................................................394.4 Tension stiffening value............................................................................................404.5 Calculation of the mean stiffness.............................................................................42

5.Beam under dead load and temperature gradient......................................445.1 M T & M cracking at both ends &T enlarge the cracking at ends and no cracking at middle span...................................................................................................46

5.1.1 Dead load effect.............................................................................................475.1.1 Temperature gradient effect.........................................................................53

5.2 Example:....................................................................................................................576.Calculation including normal force.................................................................59

6.1 Analysis procedure....................................................................................................616.1.1 Determine the compression zone height......................................................616.1.2 End rotation calculation...............................................................................626.1.3 End moment and moment due to temperature gradient...........................66

6.2 Design stiffness and mean stiffness..........................................................................67


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6.3 Comparison of mean stiffness with different situations.........................................706.3.1 Comparison with different temperature gradient......................................706.3.2 Comparison with different normal force.....................................................756.3.3 Comparison with different q load................................................................80

7.Bio-diesel project............................................................................................................847.1 Project analysis..........................................................................................................857.2 Redesign of the project.............................................................................................86

8.Program for obtaining mean stiffness..............................................................909.Conclusion and recommendations.......................................................................94References......................................................................................................................................96Appendix 1.....................................................................................................................................97Appendix 2...................................................................................................................................101Appendix 3...................................................................................................................................114Appendix 4...................................................................................................................................115Appendix 5...................................................................................................................................130Appendix 6...................................................................................................................................140

Predictingofthestiffnessofcrackedreinforcedconcretestructure YongzhenLi

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Chapter 1

General introduction


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1.1 Introduction Nowadays, concrete is one of the most important construction materials in the world. Concrete projects are distributed in many fields, such as buildings, tunnels, bridges and so on. Concrete is a kind of construction material with high compressive strength and a good durability, but a relatively low tensile strength. The tensile strength of concrete is much lower than its compressive strength. Cracks might occur in concrete at a low tensile stress. Cracking is inherent in design of reinforced concrete. These cracks might influence the structures durability and its appearance. If the cracks are too wide, the structure might not fulfill requirements with regard to durability and service ability e.g. liquid tightness. Therefore a good design and detailing of a structure should be made to limit crack widths.

Fig. 1-1 Cracks in a concrete structure

1.2 Problem description

In order to prevent the failure of a structure caused by cracking, a good understanding of cracking is required. Usually, cracks which have small width will not or hardly affect the structure. The crack width should therefore be controlled under a limit level. Unexpected or excessive cracking might occur. An example is the new cast wall which is restrained at both sides at early age when there is no external loading on it. But still some cracks might occur as shown in Fig.1-2. The design of the wall is ok with regard to ULS design, but why are there some cracks? What is the reason for the formation of these cracks? Might these cracks influence the durability of the structure? How to model these cracks with the cracks together caused by the other actions?


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Fig.1-2 Early age cracking on concrete structure

Generally, cracking can be caused by various kinds of reasons, such as external loading, restrained deformation, creep and so on. External loading and restrained deformation always are the main reasons of cracking. But codes often deal extensively with the first category which is the external loading, whereas the second category is hardly dealt with. Therefore, cracking caused by restrained deformation might be ignored by using codes to design. The cracking in the Fig.1-2 mostly is caused by the restrained deformation. Or cracks might be caused by a combination of external loading and restrained deformation. The key point is how to calculate the stiffness, for the stiffness is used to transfer a restrained deformation from an action to an effect on the structure. Before cracking, the stiffness will be constant as the stiffness of the uncracked cross-section, but if the concrete is cracked by external loading or restrained deformation, the stiffness of the structure also change. The stiffness will decrease as the cracking increases.

1.3 Goal of the research The goal of this research is to find an expression of the structural crack width calculation for cracking caused by different action combinations in different structures, such as combinations of external loading and thermal deformation, or external moment and imposed curvature. There might be a difference in the order of the actions that occurred. So what is the difference between the imposed deformation first and the external loading first? Is there also any difference when an imposed deformation and an external load occur together? Firstly cracking caused by an individual action should be investigated. After this, there is a problem about how to combine the individual actions. In order to solve the problem of action combination, the stiffness of the structure should be calculated exactly. Also the conversion from the structure action to the cross-sectional effect is another important point.


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1.4 Research outline

Fig. 1-3 Outline of the research


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Chapter 2

Literatures survey


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In the literatures survey chapter ,the main contents will be presented as two parts: crack width calculation in different codes and theoretical model for the calculation the crack width. The literature study on comparison with different codes includes four different codes to calculate the crack width.

Dutch Code Eurocode American Code Chinese Code

The literature study on theoretical models is divided into four parts:

Introduction of crack width control Causes of crack formation Crack combinations Height of compression zone Continuous theory to determine crack widths

2.1 Different codes in calculation crack

Nowadays, in an actual project, the crack width has always been calculated by following a Code. But there are differences between different codes, which depend on their different theories. Four different codes will be compared, namely the Dutch Code, EuroCode, American Code and Chinese Code. In these codes, the theories for the calculation of crack width are not totally the same. Mostly, the equations are found by an empirical equation or a semi-theoretical and semi-empirical equation. Crack width control based on steel stress and bar diameter/spacing is derived from crack width equations, so their basis is the same in different codes. But the calculation methods or criteria in different code have a little difference. For example, in the American code the crack width will be controlled by controlling the reinforcement stress or bar spacing. In the Chinese code it will be controlled by calculating the crack width and comparing with the maximum width. In the Eurocode both methods are mentioned. For these two methods, the basic theory is the same. If the maximum crack width is substituted into the equation of the crack width calculation, the maximum steel stress will be obtained. So crack width control might be transferred into steel stress control which is much easier for an engineer to use. 2.1.1 Crack width calculation equations and comparison

In the Eurocode 1992-1-1 and the Chinese Code GB50009 is presented the method of directly calculating the crack width to control crack width. The equations to calculate the crack width are shown below.


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In the Eurocode 1992-1-1 [4], the equation of calculating crack width is,

,max ( )k r sm cmw s (1-1)

,maxrs is the maximum crack spacing ,max 3 1 2 4 ,/r p effs k c k k k Eq.(7.11) in [4]

sm is the mean strain in the reinforcement under the relevant combination of loads, including the effect of imposed deformations and taking into account the effects of tension stiffening. Only the additional tensile strain beyond the state of zero strain of the concrete at the same level is considered

cm is the mean strain in the concrete between cracks Where

,,

,

(1 )0.6

ct effs t e p eff

p eff ssm cm

s s

fk

E E

(1-2)

s is the stress in the tension reinforcement assuming a cracked section. For pretensioned members, s may be replaced by p the stress variation in prestressing tendons from the state of zero of the concrete at the same level.

e is the ratio /s cmE E

,p eff is 2 '

1

,

( )s pc eff

A AA

1k is a factor dependent on the duration of the load

,c effA is the effective area of concrete in tension surrounding the reinforcement or

prestressing tendons of depth, see Fig 2-1

Fig 2-1 Effective tension area of cross section


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And in the Chinese Code GB50009 [7], the equation for calculating crack width is

max (1.9 0.08 )eqsk

crs ts

dw c

E (1-3)

(1-4)

cr is the coefficient in [7] table 8.1.2-1 is a strain coefficient of steel between cracks. When 1, then =1.

1.1 0.65 tkte sk

f

sk is calculated in [7] equation 8.1.3

tkf is the concrete tensile strength.

c is concrete cover, when c65mm, then c=65mm.

te is the ratio of reinforcement in the effective tension zone which is similar as

,c effA in Fig 2-1. When te


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the number of bars or wires . s is bar spacing.

In Eurocode 1992-1-1 [4] and the Chinese Code GB50009 [7], the allowable crack width will be determined by different exposure class and reinforcement condition. The allowable crack width is derived from Table 1 and Table 2 in Eurocode 1992-1-1 and Chinese Code GB50009. The maximum crack width in Eurocode will be found in Table 2-1 [4].

Table 2-1 Recommended values of wmax in Eurocode 1992-1-1 [4] In Table 2-1, the exposure class is defined in Table 2-3 as below:


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Table 2-3 Exposure class in Eurocode 1992-1-1


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And the maximum crack width in Chinese Code GB50009 will be found in Table 2-3.

Exposure

Class

Only reinforced members in concrete

Prestressed members in the concrete

Cracking control level

Crack width (mm)

Cracking control level

Crack width (mm)

1 3 0.3 3 0.2 2 3 0.2 2 Decompression3 3 0.2 1 DecompressionTable 2-3 Recommended values of wmax in Chinese Code GB50010 [7]

In Table 2-3, the exposure class is defined as below: Exposure Class 1: Normal environment indoor. Exposure Class 2: Moist environment indoor or outdoor except in cold area and corrosion environment. Exposure Class 3: The other exposure condition. Compared with the above two tables, the Chinese code GB50009 seems more strictly than Eurocode 1992-1-1. And in Eurocode, it is divided with bonded tendons. But in Chinese Code GB50009, it is divided with whether contain pre-stressed reinforcement. The crack width calculation equations in the codes and its influential factors are compared in the following Table 2-4. illustrates that the factor is present the equation. illustrates that the factor is not in the equation but it is already considered in the equation.

Influence factor Direct method in Eurocode 1992-1-1

Chinese Code GB90005

Eq.(5) refer to ACI 318-02

Concrete cover thickness

Concrete tensile strength

E modulus of steel Steel stress

Reinforcement diameter

Bar spacing Exposure

environment

Tension reinforcement ratio

Effective tension area of concrete

Table 2-4 Comparison the factors in crack width calculation equations of three codes


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From Table 2-4 and above equations, some conclusions will be obtained as below. 1> All the equations consider the concrete cover thickness, steel stress, steel

diameter and bar spacing. Especially for the steel stress which directly influences the crack width.

2> Changing the steel diameter, bar spacing and reinforcement ratio has impact on the crack width. It will also respond to a change in the steel stress. So in American code ACI 318-02 and Dutch Code NEN6720, they use the method of controlling steel stress to control crack width is used.

3> This direct calculation method is more complicated compared with the other methods. When changing the reinforcement properties, it should be recalculated again.

2.1.2 Steel stress equations under crack width control and comparison

By using this method, it is only necessary to substitute the structural parameters and exposure parameters in to the equations to find out the maximum allowable steel stress. And comparing the steel stress with the maximum steel stress, it will let the engineers know whether it is sufficient. Or even it can use the steel stress and the maximum crack width to determine the maximum bar spacing or bar size. The following tables will illustrate the different indirect crack width controls in different codes.

Table 2-5 Maximum bar diameters for crack control in Eurocode 1992-1-1 [4]


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Table 2-6 maximum bar spacing for crack control in Eurocode 1992-1-1 [4]

Table 2-7 maximum bar spacing for crack control in ACI 318-02 [5]

Table 2-8 Maximum bar diameters for crack control in NEN6720 [6]

2.1.3 Conclusion

From above it can be seen that there are two different methods to compute crack width. One method is to calculate the crack width directly and compare with the maximum width. On the other hand, detailing requirements with regard to bar diameter or bar spacing linked to steel stress are linked with the crack width equation, acquired by presenting this equation in a different form. The latter method is more convenient for engineers, which do not need to calculate the crack width.


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2.2 General literatures 2.2.1 Introduction of crack width control

Generally, a concrete crack is generated when the stress in the concrete is larger than the cracking stress. So in a reinforced concrete cross-section the concrete carries the compressive stress and the reinforcement has to carry the tensile stress. In the beginning of the crack stage, if too little reinforcement is used, the crack can be too wide, even if the cracking force is only exceeded to a small extend. So we also have to define a maximum crack width value to check whether the crack in the structure is sufficient.

2.2.2 Causes of cracks Though there are many reasons for cracking, the main reasons are external loading and restrained deformation.

2.2.2.1 External Loading

From [1, 2], the axial force strain diagram of a reinforced concrete tension member is obtained see Fig.2-1. From this diagram it can be seen that there are four stages of cracking behavior.

Fig. 2-1 The axial force strain relation diagram in a reinforced concrete

Stage I is the uncracked stage. In this stage concrete does not crack, and the axial


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force is smaller than the cracking force of the concrete Ncr. So in this stage, the equivalent stiffness is equal to the concrete stiffness.

E ( )b m CSA EA (2-1) Eb is the concrete elastic modulus

mA is the equivalent area of the section which is transfer the steel area into

concrete area by times sc

EE

.

Stage II is the crack development stage. This stage only occurs under the condition of fdc>>cr and N = Ncr. When the imposed strain is larger than the cracking strain, the crack will occur, and in the whole stage II the axial force will be equal to cracking force. So in this stage, the equivalent stiffness will be computed by the following equation,

E crb mm

NA (2-2) m is the mean steel strain

Stage III is the crack widening stage and the crack pattern is fully developed. In this stage, the number of cracks will be constant while their width will increase. The tensile force will be fully carried by the steel, and the bond force will transfer part of the force from the steel to the concrete. In this stage, the equivalent stiffness will be calculated by the following equation,

( )E IImb mm

A EA

(2-3)

IIEA is the steel stiffness only.

m is the mean steel strain.

ts is the tension stiffening. Stage IV is the final stage. In this stage, the force reaches the yield strength of the steel. The deformation will increase when the force remains unchanged. The equivalent stiffness will be calculated by the following equation,

E syb mm

NA (2-4)


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2.2.2.2 Imposed strain

Mostly the response caused by an imposed strain is similar to the axial force without two major differences. Firstly, the force caused by an imposed deformation dose not exceed the stiffness of the tensile member in stage I times the imposed strain,

( )csN EA (2-5) This stage it only present for small imposed deformations, because the imposed deformation must be limited to

cr (2-6)

cr is cracking strain of concrete So mostly ( )csEA will be larger than N . On the other hand, the length of the stage II largely depend on the reinforcement ratio. Because the external loading is constant in this stage, so the lower reinforcement ratio will cause the longer in this stage.

2.2.3 Loading combination In actual projects, there usually is not only one action that will act on the structure. External loading and deformation will often take place together. From the previous part the substantive force or deformation calculation method and theory are obtained. But if two or more different types of force and deformation together are combined, what will happen? From [1], several examples will be demonstrated here.

2.2.3.1 Axial force and Imposed strain in tensile member 1> The external load will occur before the restrained deformation.

crN N : If the external load is larger than the cracking force, then the crack pattern will be fully developed which directly in the 3rd stage in Fig.4. After that, the imposed deformation will be added. The existing crack will become larger due to the imposed deformation. From [1], to calculation of crack width, there is not a purely theoretically exact method. In this method, the incremental crack width is


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the mean crack spacing times the imposed strain at the level of reinforcement,

which is also equivalent to the steel stress of s vE .

crN N On the other hand, if the external load will not exceed the cracking force, the member will be in the 1st stage in Fig.2-1. The crack is caused by the following imposed strain. This is mostly a not fully developed crack pattern, since a fully developed crack pattern will be found for a really large imposed strain. 2> The external load will occur after the imposed deformation.

fdc If the imposed deformation is larger than the fully developed crack pattern strainfdc. , an increased of the external load will result in an increase of the stress in the steel. So the increase of the steel stress is /F sN A . So the total

stress , ( ) Fs s cr crs

Nat NA

.

fdc The crack pattern now is not fully developed due to the imposed deformation. So the following external load will cause the fully developed crack pattern mostly. In [1], the calculation method can be that the resulting steel stress in a crack is: if the

concrete is not cracked under imposed deformation ( ) /c F sE N A ; if the concrete is cracked under imposed deformation ( ) /cr F sN N A

2.2.3.2 Bending moment and imposed strain 1> Bending moment before imposed strain If the bending moment is larger than the crack bending moment, there will be a fully developed crack pattern. This is similar as in section the 2.2.3.1, the widening of existing cracks will be caused by the imposed strain. If the bending moment is smaller than the cracking bending moment, there will be a not fully developed crack pattern. So in [1], the fully developed crack pattern will mostly develop, and the resulting steel stress is the steel stress due to


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the bending moment plus the stress due to the imposed deformation times the

incremental stiffness from stage III / ( )q s s vM zA E . 2> Bending moment after imposed strain If the imposed strain is larger than the cracking strain, the bending moment

causes an increase of the steel stress equal to / ( )q sM zA . The stress due to the

strain will be calculated by using Fig.2-1. On the other hand, if the imposed strain is smaller than the cracking strain,the crack pattern will stay in the not fully developed pattern. From [1], the resulting

steel stress in a crack is , / ( )s cr q sM zA From the above we can obtain that a critical part in a loading combination is the definition of the stiffness. In general, the stiffness will be estimated in practice. But this value is often not exactly correct. Considering this, we will work out a more exact result in the latter part of the thesis.


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Chapter 3

Calculation of compression zone

height


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In this section, the calculation of compression zone height in different condition and with different forces will be illustrated.

Fig 3-1 Compression zone height in a concrete element The compression zone is the compressive area of the cross section after cracking. It will bear the compressive stress due to moment or a normal force. Also in the cross section equilibrium, the compressive force taken by the compression zone is equal to the tensile force taken by tension reinforcement for pure bending. If the compression zone height is equal to zero, that means the cross section is all under tensile stress. On the other hand, if the compression zone height reaches its maximum value which is equal to h, the section is all under a compressive stress. The compression zone height is a very important parameter in the concrete cross section calculation. The stiffness of the cross section after cracking largely depends on the compression height. So obtain the exact value of the compression zone height is necessary. In order to obtain the compression zone height in a crack, we need to calculate in a cracking cross section as Fig 3-2. The Normal force balance 0N and moment balance

0M will be used in the calculation for solving the compression zone height. And the cross section parameters and materials parameters also are needed for solving the compression zone height. The calculation process will be found in the following sections and Appendix.1. The compression zone height calculation will be divided into several conditions as below: 1> Only moment & only tension reinforcement (bottom reinforcement) 2> Moment and Normal force & only tension reinforcement (bottom reinforcement) 3> Only moment & both compression and tension reinforcement 4> Moment and Normal force & both compression and tension reinforcement At last, the general equation for solving the compression zone height will be obtained.


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Fig 3-2 Cross section in a crack 3.1 Compression zone height under only tension reinforcement and only

bending moment In this condition, there is only bending moment and tension reinforcement. So the cross section parameters are shown in Fig 3-3 as below.

Fig 3-3 Cross section M & only tension reinforcement

By using the normal force 0N , the compression zone height will be calculated. The stress and strain distribution is shown in Fig 3-4. The process of obtaining the


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compression zone height is illustrated later.

Fig 3-4 Stress and strain of the cross section

By using the normal force equilibrium 0N , which is derived from the stress distribution in Fig 3-4 the following relationship will be obtained:

0 0c sN N N Where,

The concrete area force 2c c cbxN E

The tension reinforcement force s s s sN A E So the relationship can be rewritten as

02 c c s s sbx E A E (3-1)

In this equation, b is the cross section width x is the compression zone height

sA is the tension reinforcement area

cE is the elastic modulus of concrete

sE is the elastic modulus of tension reinforcement

c is the concrete strain

s is the tension reinforcement strain


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The relationship between concrete strain c and reinforcement strain s can be derived by the strain relation in the Fig 3-4 as below

s cd xx

(3-2)

If we substitute Eq. (3-2) into Eq. (3-1), the following relationship will be obtained:

( ) 0 02 2c c s s c s sbx d x bx d xE A E E A E

x x (3-3)

Finally, the compression zone height x will be obtained by solving the Eq. (3-3) and the result is as below.

( 2 )s s s s s s cc

A E A E A E E bdx

E b (3-4)

If we use the ratio of E-modulus sec

EE

and reinforcement ratio sAbd

in the

equations, the Eq. (3-4) will be rewritten as below.

2( ( ) 2 )e e ex d (3-5)

3.2 Compression zone height under both compression and tension reinforcement with only bending moment In this condition, there is only bending moment with both compression and tension reinforcement. So the cross section basic parameters are shown in Fig 3-5 as below.

Fig 3-5 Cross section M & both compression and tension reinforcement


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By using the normal force 0N , the compression zone height will be calculated. The stress and strain distribution is shown in Fig 3-6. The process of obtaining the compression zone height is illustrated later.


By using the normal force equilibrium 0N , which is derived from the stress distribution in Fig 3-8 the following relationship will be obtained:

0 0c scomp sN N N N Where,


The tension reinforcement force s s s sN A E The compression reinforcement force scomp scomp scomp scompN A E So the relationship can be rewritten as

02 c c scomp scomp scomp s s sbx E A E A E (3-6)



scompA is the compression reinforcement area


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scompE is the elastic modulus of compression reinforcement



scomp is the compression reinforcement strain

The relationship between concrete strain c , tension reinforcement strain s and compression reinforcement strain scomp can be derived by the strain relation in the Fig 3-6 as below

s cd xx

(3-7) u

scomp cx cx

(3-8) If we substitute Eq. (3-7) and Eq. (3-8) into Eq. (3-6), the following relationship will be obtained:

( ) 02

02

uc c scomp scomp s s

uc scomp scomp s s

x cbx d xE A E A Ex x

x cbx d xE A E A Ex x

(3-9)

Finally, the compression zone height x will be obtained by solving the Eq. (3-9) and the result is as below.

2 2 2 22 2 2s s scomp scomp s s s scomp s scomp c s s scomp scomp c u scomp scompc

A E A E A E A A E E E bdA E A E E bc A Ex

E b

(3-10)


EE

, scompescompc

EE

and reinforcement

ratio sAbd

, scompscomp Abd in the equations, the Eq. (3-10) will be rewritten as below.

2 2( ( ) 2 ( ) 2 2 )e escomp scomp e e escomp scomp escomp scomp escomp scomp ex d


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(3-11) 3.3 Compression zone height under only tension reinforcement with both

bending moment and normal force In this condition, there is both normal force and bending moment with only tension reinforcement. So the cross section basic parameters are shown in Fig 3-7 as below.

Fig 3-7 Cross section M and N & only tension reinforcement

By using the normal force 0N and 0M , the compression zone height will be calculated. The stress and strain distribution is shown in Fig 3-8. The process of obtaining the compression zone height is illustrated below.



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By using the normal force equilibrium 0N and 0M , which are derived from the stress distribution in Fig 3-8 the following relationship will be obtained:

0 s cN N N N 0 ( ) ( )

2 2 3s ch h xM N d N M

Where,


The tension reinforcement force s s s sN A E h is the total height of the cross section d is the distance between the reinforcement and the top of the cross section So the relationship can be rewritten as

2s s s c cbxA E E N (3-12)

( ) ( )2 2 3 2c c s s sbx h x hE A E d M (3-13)








s cd xx



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( )2c s s c

d x bxA E E Nx

(3-15) ( ( ) ( ))

2 2 3 2c c s sbx h x d x hE A E d M

x (3-16)

From equations (3-15) and (3-16), the relationship between M/N ration and compression zone height x is as below:

( ) ( )2 2 3 2

2

c s s

s s c

bx h x d x hE A E d Mxd x bx NA E Ex

(3-17)

So we can obtain the value of compression zone height x by given a certain /M N value.


EE

, the moment and normal force ratio MeN

and reinforcement ratio sAbd

, in the equations, the Eq. (3-17) will be rewritten as below.

( ) ( )2 2 2 2

2

e

e

x h x d x hdd x ed x x

x d

(3-18)

3.4 Compression zone height under both compression and tension reinforcement with both normal force and bending moment In this condition, there is both normal force and bending moment with both compression and tension reinforcement. So the cross section basic parameters are shown in Fig 3-9 as below.


29

Fig 3-9 Cross section N and M & both compression and tension reinforcement

By using the normal force 0N and 0M , the compression zone height will be calculated. The stress and strain distribution is shown in Fig 3-10. The process of obtaining the compression zone height is illustrated below.


By using the normal force equilibrium 0N and 0M , which are derived from the stress distribution in Fig 3-10 the following relationship will be obtained:

0 s c scompN N N N N 0 ( ) ( ) ( )

2 2 2 3s scomp u ch h h xM N d N c N M


30

Where,


The tension reinforcement force s s s sN A E The compression reinforcement force scomp scomp scomp scompN A E h is the total height of the cross section d is the distance between the reinforcement and the top of the cross section So the relationship can be rewritten as

2s s s c c scomp scomp scompbxA E E A E N

(3-19)

( ) ( ) ( )2 2 3 2 2c c scomp scomp scomp u s s sbx h x h hE A E c A E d M (3-20)



scompA is the compression reinforcement area



scompE is the elastic modulus of compression reinforcement



scomp is the compression reinforcement strain


s cd xx

(3-22)


31

uscomp c

x cx


( )2

uc s s c scomp scomp

x cd x bxA E E A E Nx x

(3-22) ( ( ) ( ) ( ))

2 2 3 2 2u

c c scomp scomp u s sx cbx h x h d x hE A E c A E d Mx x

(2-23) Finally, the compression zone height x will be obtained by solving the Eq. (3-9) and the result is as below.

( )( ) ( ) ( )( )2 2 2 3 2

( ) ( )2

us s c scomp scomp u

uscomp scomp s s c

x cd x h bx h x hA E d E A E c Mx xx c d x bx NA E A E Ex x

(3-24)


EE

, scompescompc

EE

and reinforcement

ratio sAbd

, scompscomp Abd in the equations, the Eq. (3-10) will be rewritten as below.

( )( ) ( ) ( )( )2 2 2 2 3 2

( ) ( )2

ue escomp scomp u

uescomp scomp e

x cd x h x x h x h cx d x ex c d x x

x x d

(3-25)


32

3.5 The compression zone height equation of Noakowski

From [17], Noakowski obtained an equation of calculation compression zone height. In his equation he moved the centroidal axis to the position where the area moment equal to 0 after cracked which is shown in Fig 3-11. His basic theory is using the area inertia equal to 0 and the stress at compression zone height position equal to 0 to obtain the relationship between compression zone height and force.

Fig 3-11 Stress and strain in situation I and II by Noakowski

In Fig 3-11, situation I is the uncracked stage, the centroidal axis is at position2h

. After

moved, the centroidal axis from o to 'o . The position of 'o is the stress equal to 0

only under moved bending moment after cracked. After calculation [17], the compression zone height equation is shown as Eq. (3-26).

4 3 2 2

23 2 2

4 12 12 12 126 18 12( ) 12

x x x x

x x x

k Ak Bk Ck AC Bk Ak A B k AB

(3-26)

In this equation,

(e s scompA (e s u scompB k

2(e s u scompC k

e is the modulus ratio between reinforcement and concrete sec

EE


33

s is the area ratio between tension reinforcement and concrete ,

ss

c eff

AA

scomp is the area ratio between compression reinforcement and concrete,

scompscomp

c eff

AA

Where ,c effA bd

uk is the related height from the reinforcement to bottom uuckd

2 is the moment and normal force ratio after moved the centroidal axis, 2

2 1 0(2 )2( )

xx

x

B kkA k

1 is the moment and normal force ratio before moved the centroidal axis,

1e Md dN

0xk is the ratio of centroidal position and total height, in rectangular cross section, it is

always 0.5.


34

3.6 Example by using two method of calculation compression zone height There is a cross section shown in Fig 3-12. The external normal force and moment is applied on it. Both top and bottom have the reinforcement. The two methods in section 3.4 and 3.5 will be applied to calculate in this example.

Fig 3-12 Cross section of example

The basic parameters of the cross section and the external force are shown as below:

2

5 2

16000 /

2 10 /c

s

E N mmE N mm

5 22 10 /scompE N mm 12.5sc

EnE

605h mm 550d mm 55uc mm 2

, 550 400 220000c effA bd mm 21000sA mm 21000scompA mm

1000 0.00455220000s

1000 0.00455

220000scomp

1434.4Me mm

N

1 2.608

MdN

Substituting the above values into the Eq. (3-25) and Eq. (3-26), the compression zone height will be obtained. The process of calculation is shown in Appendix 1. After calculation, the results of Noakowski equation and Eq. (3-25) are similar which are


35

165mm and 164.99mm respectively. The difference between Eq. (3-26) and Eq. (3-25) is that, for Eq. (3-25) the basic theory is the cross section strain and stress balance. But for Eq. (3-26), Noakowski use the area moment equal to 0 and stress balance after moved centroidal axis to obtain the compression zone height equation. The result obtained by both methods is similar, so that either of the methods can be used in compression zone height calculation. The 1 -kx curves by using equation (3-25) and (3-26) are plotted as Fig 3-13 and Fig 3-14. In these figures, 1 is the moment and normal force ratio, and kx is related height of compression zone height.

Fig 3-13 1 -kx relationship curve by using Eq. (3-25)


36

Fig 3-14 1 -kx relationship curve by using Eq. (3-26)

Chapter 4

Calculation of the stiffness


37

Stiffness is a very important parameter in crack width calculation. Always the stiffness can be easily found if only one type of action is applied on the element. But if there are two or more different actions on the structure, such as external loading and restrained deformation, the calculation of stiffness is much more complicated.

For obtaining the mean stiffness of the element ( )mEI and ( )mEA , the Tension Stiffening

Law [2, 3, 16, 17] is applied. The moment-curvature relationship in Fig 4-1 is used to obtain a realistic determination of the bending moment or normal force. Also after having obtained the stiffness of the element, every parameters of the element in both the uncracked and the cracked stage will be known. So the mean stiffness of the element is the key parameter in the future calculation.

Fig 4-1 Tension Stiffening Law moment-curvature relationship

By using Tension Stiffening Law in Fig 4-1, the stiffness calculation process is divided into several steps in Fig 4-2 as below, and it will be described detailed in the following sections.


38

Fig 4-2 Process of mean stiffness calculation

4.1 Cracking Force

In the cracking force calculation, there are three conditions. For every condition, there is a different expression for the cracking force. 1> Only bending:

Cracking moment cr ctM W f W is the area moment of the cross section in the uncracked stage

ctf is the concrete cracking tensile strength

2> Only Normal force

Cracking normal force cr ctN A f A is the area of the cross section in the uncracked stage

3> Both bending and normal force

Cracking moment with a constant normal force 0cr ctNM W fA

0oN

for tension.


39

4.2 Bending stiffness in a crack

From Fig 4-1, the stiffness in a crack ,( )s crEI is equal to the mean stiffness of the

element in stage II.

,( )II

s cr cEI E I (4-1) And the inertia modulus in a crack is equal to

32 2( ) ( )

12 2II

e sbx xI bx A d x (4-2)

b is the cross section width x is the compression zone height


e is the ratio of E-modulus where sec

EE

So that the stiffness in a crack should be

32 2

,( ) ( ( ) ( ) )12 2II

s cr c c e sbx xEI E I E bx A d x

(4-3)

The stiffness in a crack can also be calculated by another method as below 1( )( )3

IIc s sE I E A d x d x

(4-4)

By using the same theory, the axial stiffness also will be obtained as below.

,( )II

s cr c cEA E A E bx 4.3 Difference of the centroidal axis x after moved

The centroidal axis moves from the original axis which always is at position 2h

in the

uncracked stage to the area moment equal to 0 after cracking which is shown in Fig 3-11. By using the area moment equal to 0 to determine the position of the centroidal axis after cracking. The difference x is

2202(

II do

d

B kW xA k

(4-5)

Where

(e s scompA (e s u scompB k


40

e is the modulus ratio between reinforcement and concrete sec

EE

s is the area ratio between tension reinforcement and concrete ,

ss

c eff

AA

scomp is the area ratio between the compression reinforcement and the

concrete,

scompscomp

c eff

AA

Where ,c effA bd

dk is the related height of compression zone dxkd

4.4 Tension stiffening value

From Fig 4-1, the distance ( ) between the mean moment-curvature line and the cracked cross section moment-curvature line is called the tension stiffening effect.

The value of ( ) describes the magnitude of the tension stiffening effect. Between the cracks the bond stresses are active and the concrete takes over part of the tension from the steel. Thus the reinforcement is being stiffened by the concrete. [16, 17]. Fig 4-3 shows the strains for calculating the crack spacing and the average strains. [18]

Fig 4-3 Strains for calculating the crack spacing and the average strains

By using the mean bond law from the Model Code [18], Fig 4-4 and Fig 4-5 show the


41

basic theory to obtain the tension stiffening value. In Fig 4-4, 1 means the maximum curvature in a cracked section at the beginning of the crack formation stage. By using the mean bond law [18], the mean curvature in a cracked section should be

1 10.44m by using the theory in CEB-FIP Model Code [18]. So that the

1 1(1 0.44) 0.56 . And the transfer length is 12a

. In Fig 4-5, the element is

in fully developed crack pattern. So the transfer length will be changed to

10.752 2

aa [2]. So the tension stiffening factor also will be changed.

Fig.4-4 Curvature at crack section in beginning of crack formation stage by mean bond

law

Fig 4-5 Curvature at crack in fully developed crack pattern by mean bond law

The value of the tension stiffening factor at a fully developed crack pattern is equal to

. 1 ,0.75 0.42s cr fdc s cr (4-6) . 1 ,0.75 0.42s cr fdc s cr (4-7)


42

Where

,s cr

is the curvature at a cracked section ,,( )

s crs cr

M N xEI

,s cr

is the strain at a cracked section ,,( )

s crs cr

NEA

4.5 Calculation of the mean stiffness

In the calculation of the mean stiffness of a element, if there is only one action on the element it is easily obtained by the tension stiffening law [1, 2, 16, 17]. But if there is two or more different actions on the element, the method should be followed as presented below. The black line is the moment-curvature effect line in total element. The red line is the moment-curvature effect in the cracked cross section. The blue line is the moment-curvature effect in the cracked cross section without M which is caused by moving the centroidal axis.

Fig 4-5 Moment-curvature relationship by applying both moment and normal force


43

The procedure to obtain the mean moment-curvature relationship of whole element which is shown as black line in Fig 4-5 is described as below:

1> Defining the crM value and position in Y axis.

2> Calculating the stiffness at a crack ,( )s crEI .

3> Using the stiffness at a crack, the slope of the blue line in Fig 4-5 can be defined. 4> Adding the extra moment due to normal force M 5> The position of red line in Fig 4-5 will be defined by using M . 6> Calculating the tension stiffening value . 7> By using tension stiffening law, the position of fdc will be found. 8> At last, the mean moment-curvature relationship of whole element shown as black

line will be obtained. From Fig 4-5, the there are 3 different stages in the cracking development: uncracked stage, crack formation stage and fully developed crack stage. So each stage will have different expression of mean stiffness.

Uncracked stage: ( )m c cEI E I (4-7)

Crack formation stage: ( )mm

MEI m s ,( )s s crM

EI

(4-8)

Fully developed crack stage: 2 1( )mm s

M M N xEI (4-9)

1M is the bending moment after moving the centroidal axis

1 , ,( )( ) ( )m s cr s s crM EI EI 2M is the original moment without moving the centroidal axis

M is the extra moment caused by moving centroidal axis M N x x is the moving distance of centroidal axis is the tension stiffening effect ,0.42 s cr

,s cr is the curvature at the cracked cross section under cracking moment

s is the curvature at the cracked cross section at the load considered

,( )s crEI is the stiffness in a crack Eq. (4-5).


44

Chapter 5

Beam under dead load and

temperature gradient


45

After discussed about the cross section problem, in this chapter, the calculation based on a simple structure will be illustrated. In order to illustrate it more clearly, a clamped beam will be discussed in this chapter to show the characteristics under different loading combination. As solving this problem, the theory of the simple element can be used in some complicated real projects. The simple element is shown in Fig 5-1. It is a clamped beam with tension reinforcement under both dead load and temperature gradient effects. The dead load can be transferred as a uniform distribute load q, and the temperature gradient is T . So there will be a bending moment due to self-weight, and also a bending moment due to thermal difference. A big problem is that how to combine these two actions together, and is there any difference among different combinations? The following calculation will illustrate it.

Fig 5-1 Clamped beam under both dead load and temperature gradient

For a combination of dead load and temperature gradient, there are many various situations. Such as dead load is given first, and then the temperature gradient is loaded, or exchange the order. For different situations, there will be different procedures and results respectively. So using an appropriate calculation procedure is essential to obtain correct results. All the different situations are listed as below:

1. M T 1.1 M Uncracked

1.1.1 T Uncracked 1.1.2 T Not fully developed cracked pattern 1.1.3 T Fully developed cracked pattern

1.2 M Cracking at ends only 1.2.1 T Ends Cracking increased

1.2.1.1 No cracking at middle span 1.2.1.2 Cracking at middle span

1.2.2 T Ends cracking decreased 1.2.2.1 No cracking at middle span 1.2.2.2 Cracking at middle span 1.2.2.3 Cracking at middle span but ends cracking disappeared

1.3 M Cracking at both ends and middle span


46

1.3.1 T Ends cracking increased 1.3.1.1 Middle span cracking is not disappeared 1.3.1.2 Middle span cracking is disappeared

1.3.2 T Ends cracking decreased 1.3.2.1 Ends cracking is not disappeared 1.3.2.2 Ends cracking is disappeared

2. T M 2.1 T Uncracked

2.1.1 M Uncracked 2.1.2 M Cracking at ends 2.1.3 M Cracking at both ends and middle span 2.1.4 M Cracking at middle span

2.2 T Not fully developed cracked pattern 2.2.1 M Cracking at ends 2.2.2 M Cracking at both middle span and ends 2.2.3 M Cracking at middle span

2.3 T Fully developed cracked pattern 2.3.1 M Cracking at ends 2.3.2 M Cracking at both middle span and ends 2.3.3 M Cracking at middle span

3. T & M 3.1 Uncracked 3.2 Cracked

3.2.1 Cracking at ends 3.2.2 Cracking at both middle span and ends 3.2.3 Cracking at middle span

In order to illustrate it, one typical condition will be discussed in the following section. And one example will be calculated to prove this theory.

5.1 M T & M cracking at both ends &T enlarge the cracking at

ends and no cracking at middle span In this situation, the moment due to uniformly distribute load q will be added first. The beam will crack at both ends for this moment. Then the temperature gradient is loaded, the cracking at ends will increase. It is shown as Fig 5-2.


47

Fig 5-2 Loading procedure So the effect caused by dead load only need to be analyzed first, then added the temperature gradient. 5.1.1 Dead load effect.

Dead load of a beam is uniformly distributed. So it can be recognized as a uniform distributed load q. The moment due to dead load should be a parabola line. In order to make the calculation easier, the linear line is instead of parabola line. The moment curve due to dead load is shown as Fig 5-3. The moment due to dead load at end is

equal to 21

12ql and at mid span is equal to 21

24ql .

Fig 5-3 Moment curve due to dead load

In the first step, the cracking length due to moment will be estimated. The length is the area where moment is larger than cracking moment on an uncracked beam which is

also shown in Fig 5-3. This estimated length is called 2L , and this value will be used in

the future calculation. Usually the estimated cracking length can be used as the area where the moment is larger than cracking moment on an uncracked beam. The cracking moment can be calculated as

216cr cr

M bh (5-1) Since the cracking length is estimated, the moment curve due to dead load will be calculated as below. In order to obtain the exact moment due to dead load, the following method will be used. Firstly, this uniformly distributed load can be divided into two parts: one is dead load on a simple support beam, and the other one is a simple support beam with a constant moment which is shown as Fig 5-4. By using this method, the clamped beam will be translated into two simple support beam problems with the rotation at ends equal to 0. It is much easier to calculate the moment curve.


48

Fig 5-4 Transition from clamped beam to simple support beam

In the Fig 5-4, 1EI is the stiffness of uncracked part, 2EI is the stiffness of cracked

part which is not fixed and ,q endM is the moment at both ends.

Considering the first part which is the dead load on the simple support beam, by using symmetry, the stiffness curve, the moment curve and the curvature curve can be easily calculated as shown in the Fig 5-5:


49

Fig 5-5 Stiffness, moment and curvature of first part

From the second graph in the Fig 5-5, the stiffness in the cracked section can be calculated which is specified in chapter 3 and 4. By applying the equation (3-4), (3-10), (3-17), and (3-14), the compression zone height of the cracked section can be drawn. In this case the compression zone height will be calculated as below,

( 2 )s s s s s s cc

A E A E A E E bdx

E b

(5-2)

After obtained the compression zone height, the stiffness on the cracked section will be calculated by using equation (4-3) as below

32 2

,( ) ( ( ) ( ) )12 2II

s cr c c e sbx xEI E I E bx A d x

(5-3)

By using the knowledge in chapter 4, the stiffness at position Lx should be calculated as below:


50

2

,

2

,

2 2( )

2 2( )

x xend DL

xx x

end DL

scr

qLL qLMEI

qLL qLM

EI

(5-4)

Where

0.42 scr

( )cr

scrscr

MEI

The final curvature at point 1, 2, 3 can be calculated as below:

21

2

LMEI

(5-5)

221

LMEI

(5-6)

31

midMEI

(5-7)

The rotation at cracked section with a certain position xL is

2

,2 2( )

x x

x crackedx

qLL qL

EI

(5-8)

And the rotation at uncracked section with a certain position xL is

2

,2 2( )

x x

x uncrackedc

qLL qL

EI

(5-9)

The rotation at end due to dead load on the simple support beam is the uncracked part plus

cracked part. So the uncracked part rotation is 1L , and the uncracked part rotation is

2L , so the total rotation at end should be , 1 2q end L L (5-10)


51

Where

22

12

2( )

L xx

L xcL

qLL qLdL

EI

(5-11)

22

220

,

2,

2

2

2( )

xL x

L xx

end DL x

xend DL x

scr

qLL qLdLqLLM qL

qLLM qL

EI

(5-12)

Considering the second part which is the constant moment on the simple support beam which is shown in Fig 5-6, by using symmetry, for the moment on the simple support beam, the stiffness curve, the moment curve and the curvature curve are shown in Fig 5-7:

Fig 5-6 The second part which is only moment applied on the beam


52

Fig 5-7 Stiffness, moment and curvature of second part From the fourth graph in Fig 5-6, the curvature at cracked and uncracked parts can be calculated as below:

,1

1

q endM

MEI

(5-13)

,2q end

Mx

MEI

(5-14)

Similar as before, the rotation at the end due to moment at this case is:

, , 1 2q endM end M M

(5-15)

Where

1 1 1M M L 2

2 20

L

M M xdL By substituting Equation (5-4), the expression of the total rotation due to moment can be written as

21

, ,20

,

2

,

1( )( )

2 2

2 2( )

L

end M x q endx xc

end DL

x xend DL

scr

L dL MqLL qLEI M

qLL qLM

EI

(5-16)

Because it is a clamped beam, there is no rotation at the end. So the rotation due to dead load plus the rotation due to constant moment should be equal to 0:

,, ,0

q endq end M end

(5-17)

Substituting (5-10) and (5-16) into (5-17), the moment at end due to dead load on a clamped beam should be

2

,,

12

0,

2

,

1( )

2 2

2 2( )

end DLq end L

xx xc

end DL

x xend DL

scr

ML dL

qLL qLEI M

qLL qLM

EI

(5-18)


53

So the moment curve due to dead load should be shown as Fig 5-8:

Fig 5-8 Moment curve due to dead load

For the cracking length 2L is estimated, so find the moment value at 2L from the

curve in the Fig 5-8.

2

22,

1( 1)2L q end

LM qL ML

(5-19)

And then compare 2L

M with crM , if they are equal, then the estimation is correct.

Otherwise, re-estimate 2L until 2L crM M . 5.1.1 Temperature gradient effect.

Caused by the dead load effect, the beam has already changed to three parts, two cracked parts and one uncracked part. The cracking length due to dead load is known

as 2L after several iterations.

The moment due to temperature gradient should be constant as the Fig 5-9. But the value should be calculated by using the rotation at end which is equal to 0. So by using the similar method in 5.1.2, the clamped beam with temperature gradient can be divided into two parts as in the Fig 5-10.

Fig 5-9 Moment due to a temperature gradient on an uncracked beam


54


In order to obtain the actual moment, the cracking length 3L also should be

estimated first. In section graph in Fig 5-10, there is a constant moment on the simple support beam. Similar with Dead load section, by using symmetry, for the moment on the simple support beam, the stiffness curve, the moment curve and the curvature curve are shown in Fig 5-11:


55

Fig 5-11 Stiffness, moment and curvature of beam

From the fourth graph in Fig 5-11, the curvature at each part can be calculated as below:

,1

2

T endMEI

(5-20)

,2

1

T endMEI

(5-21)

Similar as before, the rotation at end due to moment at this case is:

3

,

1, 1 2 ,

0

1( )( )T end

L

M end T end xc c x

LM dLE I EI

(5-22)

Where


56

2

, ,

2

, ,

2 2( )

2 2( )

x xend DL end T

xx x

end DL end T

scr

qLL qLM MEI

qLL qLM M

EI

(5-23)

In third graph in Fig 5-10, there is a rotation at end on the simple support beam. The value of this rotation can be directly calculated by the temperature gradient and temperature gradient coefficient as below:

1 3( )rotation T L L (5-24) From equation (5-24) and (5-22), the total rotation should be equal to 0 for it is a clamped beam.

, ,0

T endM end rotation (5-25)

Substitute (5-22) and (5-24) into (5-25), the moment due to temperature gradient will be obtained:

3

1 3

12

0, ,

2

, ,

( )1 )

2 2

2 2( )

TT L

xx xc c

end DL end T

x xend DL end T

scr

L LML dL

qLL qLE I M M

qLL qLM M

EI

(5-26)

At last, plus the moment due to temperature gradient and dead load, the final moment

curve will be obtained. Then the moment at position 3L should be calculated and

compare with cracking moment crM . If they are same, that means the estimated

cracking length 3L is correct. If not same, the cracking length 3L should be

estimated again until they are same.


57

5.2 Example:

There is an example which will be used by Tension Stiffening Law to calculate the exact moment distribution shown in Fig 5-12. This is a clamped beam with dead load and a constant temperature gradient. All the parameters of the beam and load are shown as below.

Fig 5-12 Clamped beam with dead load and temperature gradient

The basic parameters:

2

5 2

30000 /

2 10 /c

s

E N mmE N mm

12.5sc

EnE

23 /cr N mm

10000L mm 605h mm 550d mm 55uc mm 2

, 550 400 220000c effA bd mm 24000sA mm

4000 0.0182220000s

40 /q N mm

25T C

510 / C

By using the calculation method in 5.1, the final moment distribution will be obtained. See workings in Appendix 2. The final moment distribution is shown in the Fig 5-13 as

below. In this example, the cracking moment is equal to 82 10 Nmm . The moment at

end and middle span are 84.387 10 Nmm , 80.6135 10 Nmm respectively. The cracking length is 1400mm at both end, and there is no cracking at midspan.


58

Fig 5-13 The moment distribution on the beam [Nmm] In order to compare the moment distribution with other condition, the moment distribution under following condition will be also calculated and compared. A. Calculation with the uncracked stiffness in the whole span.

. Fig 5-14 Moment distribution with uncracked stiffness

B. Calculation with the 1/3 uncracked stiffness in whole span. [Nmm]

Fig 5-16 Moment distribution with 1/3 uncracked stiffness whole span [Nmm]


59

Chapter 6

Calculation including normal

force


60

Actually, there are many actions on a structure in real projects. Due to many complicated factors, moment and normal stress are the direct effects on the structure. All the loading actions are transformed into effects on the structure. If the complicated system can be easily analysis. In Chapter 5, the clamped beam under both temperature gradient and external loading is discussed. These actions cause moments only, whereas a normal stress will be caused by a normal force. After having applied the normal force, the reinforcement ratio and compression zone height will be changed. Also the moment distribution and stiffness will be changed as the reinforcement ratio and compression zone height changed. So if there is a normal force on the clamped beam, what will be changed? Will one third of the uncracked stiffness still be suited for designing the reinforcement correctly?

Fig 6-1 Clamped beam under temperature gradient, uniformly distributed load and normal force

In this chapter, the clamped beam will be also used, and the normal force is considered too, which is shown in Fig 6-1. With the similar method of chapter 5, the final state of the moment curve and mean stiffness will be obtained. The specific procedure is shown as Fig 6-2 as shown below.

Fig 6-2 Analysis procedure

There is another matter to which attention should be paid attention that the source of the


61

normal force. Usually the normal force is a mechanic load, but there is another important part which might be ignored easily. That is the temperature difference between construction temperature and operating temperature. Due to the difference between two states, some shrinkage or expansion will occur. It also will produce a normal stress. Thus in this chapter, the normal forces source will be considered as both mechanic load and temperature difference.

6.1 Analysis procedure The procedure is similar to the one used in chapter 5, though a little more complicated since including the normal force implies that there is one more procedure in the calculation of the compression zone height. In order to make the calculation procedure much easier, both dead load effect and temperature gradient effect will be considered together in this chapter which is different from that in chapter 5. In this way, the moment of each individual effect is not read directly, but is the result that can be calculated in one formula. The procedure is described below. 6.1.1 Determine the compression zone height

If the reinforcement is unknown, the reinforcement ratio should be determined by cross-section equilibrium.But the compression also is unknown, so the reinforcement ratio and compression zone height need to be estimated first. In the estimated process, one empirical formula is given as below which is close to the correct value.

max 0.39A0.9 0.9

Ts

s s

M N hd d

(6-1)

In this equation, the value of maxM contain both moment due to external load and temperature gradient. The moment due to temperature gradient can be calculated by using one third of the uncracked stiffness first which is an estimated value but close to the accurate value. After estimated the compression zone height and reinforcement, the following two equations should be checked. If they are not equal, the compression zone height and reinforcement should be estimated iterative again until equation (6-2) and (6-3) are tenable.

max( ) ( )( )

2 2 2 2( )

2

e

Te

x h x d x hd Md xd x x Nx d

(6-2)


62

max ( )2 3( )

3

T

s

s

h xM NA xd

(6-3)

When equation (6-2) and (6-3) are tenable, the required reinforcement sA and

compression zone height x are obtained. The two values will be used in the future calculation. With the value of compression zone height and reinforcement ratio, the stiffness at cracked cross-section will be obtained by equation (6-4) as below.

32 2( ( ( ) ( ) ))

12 2s

scr c sc

Ebx xEI E bx A d xE

(6-4)

6.1.2 End rotation calculation

The clamped beam will be divided into 3 parts which is shown in Fig 6-3: simple support beam with moment at ends; simple support beam with a rotation at end; simple support beam with uniform distributed load on the beam. The length of

cracked area will be estimated. The length of end cracked area is estimated as 2L ,

also the length of middle cracked area is estimated as 3L . These two parameters

should be given the estimated values which will be used in the future calculation. At the end of the calculation, these will be checked and iterative.



63

After estimated the length of cracked zone, the mean stiffness can be obtained by using equation (6-4) by tension stiffening law. According to section 4.5, the stiffness

at position xL will be obtained as below.

2

22 2

2 2

total x xend

xtotal x x

end

scr

qL L qLMEI

qL L qLM

EI

(6-5)

Before calculation, the moment at end endM should be estimated too. Using this

estimated moment in the calculation of rotation at end. For the uniformly distributed load on the beam, with the similar method in chapter 5, by using symmetry system the stiffness, moment and curvature distribution will be shown in Fig 6-4 as below.

Fig 6-4 Stiffness, moment and curvature distribution

From Fig 6-4, the curvature at section 1, 2 and 3 can be obtained with the stiffness and moment.


64

2

2 2total x x

xx

qL L qL

EI

(6-6)

Then the rotation at end due to the uniformly distributed load will be obtained by equation (6-6), which is shown as below: Total rotation of section 2:

2,

2

2, 20

2

2 2

2 2

2 2

estimatedtotal x xL

L DL xtotal x x

end

total x xend

scr

qL L qL

dLqL L qLM

qL L qLM

EI

(6-7)

Total rotation of section 1:

1 2,

2,

2

1,2 2

estimated

estimated

total x xL L

L DL xc cL

qL L qL

dLE I

(6-8) Total rotation of section 3:

1 2,

2

2

1, 2

2

2 2

2 2

2 2

total

estimated

L total x x

L DL xtotal x xL L

end

total x xend

scr

qL L qL

dLqL L qLM

qL L qLM

EI

(6-9)

The total rotation at end of the whole beam:

, 1, 2, 3,end DL L DL L DL L DL (6-10) For the second part which is a simply supported beam with a moment at its ends, a similar method can be used to obtain the rotation at the end. The stiffness, moment and curvature distribution are illustrated in Fig 6-5 below.


65

Fig 6-4 Stiffness, moment and curvature distribution The total rotation at end in this situation is

2,

1 2,

, 20

2

21

2

2

1(

2 2

2 2

1 )

2 2

2 2

estimated

total

estimated

L

end M end xtotal x x

end

total x xend

scrL

xtotal x xc c L L

end

total x xend

scr

M dLqL L qLM

qL L qLM

EI

L dLqL L qLE I M

qL L qLM

EI

(6-11)

For the third part which is only a rotation at the end of the simply supported beam, the rotation can be easily calculated by the temperature gradient.

, 2total

end TL

(6-12)


66

6.1.3 End moment and moment due to temperature gradient In order to obtain the moment at end, the boundary condition of the beam is needed. Because it is clamped beam, there is no rotation at end. The sum of rotation at end of all three parts should be equal to zero which is shown in equation (6-13). If it is simple support beam or one end simple support and the other end clamped, the boundary condition will be changed.

. , , 0end DL end M end (6-13) By using the boundary condition, the relationship of moment and other parameters will be obtained. Substitute equation (6-10), (6-11) and (6-12) into (6-13), the equation of moment at end will be obtained as below.

2, 1 2,

2,

2 2 2

2 20

2

2 2 2 2 2 22

2 2 2 2

2 2

estimated estimated

estimated

total x x total x x total x xL L Ltotal

T x xtotal x x total x xc cL

end end

total x xend end

scrend

qL L qL qL L qL qL L qLL dL dL

qL L qL qL L qLE IM M

qL L qL qM M

EIM

1 2,

2,

1 2,

2

2

12 2

0

2 2

2 2

1 1

2 2 2 2

2 2 2 2

total

estimated

totaestimated

estimated

L

xL L

total x x

scrL

L

xtotal x x total x xc c L L

end end

total x x total x xend end

scr scr

dL

L L qL

EI

LdLqL L qL qL L qLE IM M

qL L qL qL L qLM M

EI EI

2l

xdL

(6-14) This end moment should be compared with estimated end moment. These two values should be equal, so that the estimation is correct. If not, the end moment should be estimated again, until the two values are equal. In order to check the estimated value,

the moment at position 2L and 3L should be calculated as equation (6-15) and

(6-16) as below.

22, 2,

2 2 2total estimated estimated

L end

qL L qLM M (6-15)

22, 1 2, 1

3

( ) ( )2 2

total estimated estimatedL end

qL L L q L LM M

(6-16)

If in the estimated stage, one of 2L and 3L is equal to 0, which means the cracks

is occu

Documents

Stiffness of Cracked Rc Structure